A wave is travelling along x-axis . The disturbance at x=0 and t=0 is A/2 and is increasing , where A is amplitude of the wave. If `Y= sin (kx-omegat+phi)`, then the initial phase is `alpha pi` . Find `alpha`

Answer

To solve the problem step by step, we will analyze the given wave function and the conditions provided.

### Step 1: Write down the wave equation
The wave is described by the equation:
\[ Y = \sin(kx - \omega t + \phi) \]

### Step 2: Substitute the initial conditions
We know that at \( x = 0 \) and \( t = 0 \), the disturbance \( Y \) is given as \( \frac{A}{2} \). Thus, we can write:
\[ Y(0, 0) = \sin(0 - 0 + \phi) = \sin(\phi) \]
Setting this equal to \( \frac{A}{2} \), we have:
\[ \sin(\phi) = \frac{A}{2} \]

### Step 3: Solve for \( \phi \)
Since \( A \) is the amplitude of the wave, we can express the equation as:
\[ \sin(\phi) = \frac{1}{2} \]
The solutions for \( \phi \) that satisfy this equation are:
\[ \phi = \frac{\pi}{6} + n\pi \]
where \( n \) is any integer. However, we will consider the principal value first.

### Step 4: Determine the increasing condition
The problem states that the disturbance is increasing at \( t = 0 \) and \( x = 0 \). This means that the derivative of \( Y \) with respect to \( t \) must be positive:
\[ \frac{\partial Y}{\partial t} = -\omega \cos(kx - \omega t + \phi) \]
At \( t = 0 \) and \( x = 0 \), this becomes:
\[ \frac{\partial Y}{\partial t}(0, 0) = -\omega \cos(\phi) \]
For the disturbance to be increasing, we need:
\[ -\omega \cos(\phi) > 0 \]
This implies:
\[ \cos(\phi) < 0 \]

### Step 5: Analyze the cosine condition
From the unit circle, we know that:
- \( \cos(\phi) < 0 \) occurs in the intervals \( \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \).

Since we have \( \phi = \frac{\pi}{6} \) from the sine condition, we need to check if this value satisfies the cosine condition.

### Step 6: Find the correct value of \( \phi \)
The value \( \phi = \frac{\pi}{6} \) gives:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} > 0 \]
This does not satisfy the increasing condition. Therefore, we need to consider the next possible value:
\[ \phi = \frac{7\pi}{6} \]
This value gives:
\[ \cos\left(\frac{7\pi}{6}\right) < 0 \]
Thus, it satisfies the increasing condition.

### Step 7: Relate \( \phi \) to \( \alpha \)
The problem states that the initial phase can be expressed as:
\[ \phi = \alpha \pi \]
From our findings:
\[ \frac{7\pi}{6} = \alpha \pi \]
Dividing both sides by \( \pi \):
\[ \alpha = \frac{7}{6} \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = \frac{7}{6} \]